Question

Answer 1

\(\dfrac{2x+\sqrt{x}}{x-\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\) (ĐKXĐ: \(x\ge0;x\ne1\))

\(=\dfrac{2x+\sqrt{x}-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

Answer 2

đk : x > 0 ; x khác 1 

\(=\dfrac{2\sqrt{x}+1-\sqrt{x}}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)